Two positively charged particles m_ 1 and m_ 2 have been accelerated across the same potential difference of 200 keV as shown below. [Given mass of m_ 1 =1 amu and m_ 2 =4 amu] The deBroglie wavelength of m_ 1 will be x

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For a charged particle accelerated through potential V, the de Broglie wavelength is

\lambda = \frac{h}{\sqrt{2mVe}}

.
The wavelength is inversely proportional to

\sqrt{m}

.
Therefore:

\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{m_2}{m_1}} = \sqrt{\frac{4}{1}} = 2

.
This means

\lambda_1 = 2\lambda_2

, so x = 2 (nearest integer).

Two positively charged particles m_ 1 and m_ 2 have been...

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